Calculate the pH when 50.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Question

I cant figure out this one

Answer 1

HBrO is a weak acid. KOH is a strong base. Mixing these two will result in the formation of a buffer until the equivalence point is reached. Looking at the reaction, we have...

HBrO + KOH ==> KBrO + H2O

HBrO + OH^- ==> BrO^- + H2O .. net ionic equation

Initial moles HBrO = 20.0 ml x 1 L / 1000 ml x 0.300 mol / L = 0.006 mols HBrO

Initial moles KOH = 50.0 ml x 1 L / 1000 ml x 0.150 mol / L = 0.0075 mols KOH

NOTE that there is more KOH than HBrO, so all the HBrO will be converted to BrO^- and there will be excess OH^-. This is past the equivalence point and thus no buffer is formed.

HBrO + OH^- ==> BrO^- + H2O

0.006.......0.0075......0.........0.........Initial

-0.006...-0.006.......+0.006.............Change

0.........0.0015....0.006................Equilibrium

Final volume = 70.0 ml = 0.0700 L

Final [OH-] = 0.0015 mol / 0.07 L = 0.0214 M

Final [BrO-] = 0.006 mol / 0.07 L = 0.0857 M

To find pH, we will find TOTAL OH- then calculate pOH and finally calculate pH:

BrO- + H2O ==> HBr + OH-

Kb = [HBr][OH-] / [BrO-] and Kb = Kw/Ka = 1x10^-14 / 2.5x10^-9 = 4x10^-6

4x10^-6 = (x)(x) / 0.0875-x and assume x is small relative to 0.0875 and ignore it in denominator

4x10^-6 = x² / 0.0875

x = [OH-] = 5.92x10^-4 M

[OH-] from excess KOH = 0.0214 M

Total [OH-] = 0.02199 M

pOH = -log 0.02199 = 1.658

pH = 14 - pOH

pH = 12.342

(be sure to check all of the math)